How does a three-wire thermistor eliminate its internal resistance?
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I have a question for everyone: In a three-wire thermistor system, the signal is transmitted through an isolation barrier and converted into a current signal, which is then sent to the card. How is the internal resistance of the signal lines eliminated during this process? In my example today, the measured value of the thermistor on-site was 118.3 ohms; at the input end of the isolation barrier in the control room, it was 124.5 ohms. Yet, the temperature displayed inside was 118.3 ohms. How is this internal resistance eliminated? Is it through the isolation barrier or the card? If an XP316 card is used, is it the card that eliminates this internal resistance? If a barrier is used, does the barrier eliminate it?
It’s not about “eliminating the internal resistance of the signal lines,” but rather about removing the impact of the resistance of those signal lines on the measurement results. Refer to the schematic diagram above; by using a three-wire system, each arm of the measurement bridge has a wire resistance. The output voltage of the bridge is proportional to the ratio of the bridge arm resistances. Since the wire resistances on both bridge arms are equal, they cancel each other out, and therefore the output voltage is independent of the wire resistance.
Balance bridge method. In high school, the “balance bridge” appeared in some *problems. If you’re interested, you can calculate it using the methods available online to gain a better understanding.